Problem 1837

1. Define the Partial Sums:
Let $\{s_1, s_2, \ldots, s_{2^n}\}$ be the set of all partial sums of the natural numbers $x_1, x_2, \ldots, x_n$. By the problem's condition, all these partial sums are distinct. We can order them such that $s_1 \leq s_2 \leq \cdots \leq s_{2^n}$.

2. Condition on Differences:
Since all partial sums are distinct, for any $1 \leq k, l \leq 2^n$, we have $s_k - s_l \geq |k - l|$.

3. Sum of Partial Sums:
The sum of all partial sums is given by:
$$\sum_{k=1}^{2^n} s_k = 2^{n-1} \sum_{i=1}^n x_i$$
This is because each $x_i$ appears in exactly half of the subsets of $\{x_1, x_2, \ldots, x_n\}$.

4. Sum of Squares of Partial Sums:
The sum of the squares of all partial sums is:
$$\sum_{k=1}^{2^n} s_k^2 = 2^{n-1} \sum_{i=1}^n x_i^2 + 2^{n-1} \sum_{1 \leq i < j \leq n} x_i x_j$$
This follows from the fact that each $x_i^2$ appears in the square of each subset sum where $x_i$ is included, and each $x_i x_j$ appears in the product of each pair of subset sums where both $x_i$ and $x_j$ are included.

5. **Expression for $\sum_{i=1}^n x_i^2$:**
Using the above results, we can express $\sum_{i=1}^n x_i^2$ as:
$$\sum_{i=1}^n x_i^2 = 2 \cdot \frac{\sum_{k=1}^{2^n} s_k^2}{2^{n-1}} - \left( \frac{\sum_{k=1}^{2^n} s_k}{2^{n-1}} \right)^2$$

6. Simplifying the Expression:
Simplifying further, we get:
$$2^{2n-1} \sum_{i=1}^n x_i^2 = 2^n \sum_{k=1}^{2^n} s_k^2 - \left( \sum_{k=1}^{2^n} s_k \right)^2$$
This can be rewritten using the identity for the sum of squared differences:
$$2^{2n-1} \sum_{i=1}^n x_i^2 = \frac{1}{2} \sum_{k,l=1}^{2^n} (s_k - s_l)^2$$

7. Lower Bound on Sum of Squared Differences:
Given $s_k - s_l \geq |k - l|$, we have:
$$\sum_{k,l=1}^{2^n} (s_k - s_l)^2 \geq \sum_{k,l=1}^{2^n} (k - l)^2$$
The sum of squared differences for integers $k$ and $l$ from $1$ to $2^n$ is:
$$\sum_{k,l=1}^{2^n} (k - l)^2 = 2^n \sum_{k=1}^{2^n} k^2 - \left( \sum_{k=1}^{2^n} k \right)^2$$

8. Calculating the Sums:
Using the formulas for the sum of the first $m$ natural numbers and the sum of their squares:
$$\sum_{k=1}^{2^n} k = \frac{2^n (2^n + 1)}{2} = 2^{n-1} (2^n + 1)$$
$$\sum_{k=1}^{2^n} k^2 = \frac{2^n (2^n + 1) (2^{n+1} + 1)}{6}$$

9. Substituting Back:
Substituting these into the inequality, we get:
$$2^n \sum_{k=1}^{2^n} k^2 - \left( \sum_{k=1}^{2^n} k \right)^2 = 2^n \cdot \frac{2^n (2^n + 1) (2^{n+1} + 1)}{6} - \left( 2^{n-1} (2^n + 1) \right)^2$$
Simplifying this expression, we find:
$$2^n \cdot \frac{2^n (2^n + 1) (2^{n+1} + 1)}{6} - 2^{2n-2} (2^n + 1)^2 = 2^{2n-1} \cdot \frac{4^n - 1}{3}$$

10. Final Inequality:
Therefore, we have:
$$2^{2n-1} \sum_{i=1}^n x_i^2 \geq 2^{2n-1} \cdot \frac{4^n - 1}{3}$$
Dividing both sides by $2^{2n-1}$, we get:
$$\sum_{i=1}^n x_i^2 \geq \frac{4^n - 1}{3}$$

The final answer is $\boxed{\sum_{i=1}^n x_i^2 \geq \frac{4^n - 1}{3}}$